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Question Number 187828 by Rupesh123 last updated on 22/Feb/23

Answered by HeferH last updated on 22/Feb/23

Commented by HeferH last updated on 22/Feb/23

i. 3α + β +ω = 180° ii. ω + α = 60° iii. α= 60° − ω 3(60°−ω) + β +ω = 180° 180° −3ω + β +ω = 180° β = 2ω ∠ACB = 3ω ∠ACE = ω ∠ACB = 3∙∠ACE ✓

i.3α+β+ω=180°ii.ω+α=60°iii.α=60°−ω3(60°−ω)+β+ω=180°180°−3ω+β+ω=180°β=2ω∠ACB=3ω∠ACE=ω∠ACB=3⋅∠ACE✓

Commented by Rupesh123 last updated on 23/Feb/23

Very nice approach, sir!

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